leetcode 226. Invert Binary Tree

Invert a binary tree.

     4
   /   
  2     7
 /    / 
1   3 6   9

to

     4
   /   
  7     2
 /    / 
9   6 3   1

解法1:

本质是输的先序遍历

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        # invert self.left
        # invert self.right
        if not root: return None
        root.left, root.right = root.right, root.left
        self.invertTree(root.left)
        self.invertTree(root.right)
        return root        

解法2，上述DFS的迭代解法：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """        
        if not root: return None
        stack = [root]
        while stack:
            node = stack.pop()
            node.left, node.right = node.right, node.left
            if node.right:
                stack.append(node.right)
            if node.left:
                stack.append(node.left)
        return root                    

解法3，使用BFS：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """        
        if not root: return None
        nodes = [root]
        while nodes:
            nodes2 = []
            for node in nodes:
                node.left, node.right = node.right, node.left
                if node.left: nodes2.append(node.left)
                if node.right: nodes2.append(node.right)
            nodes = nodes2
        return root